Is sec(α)-cos(α)=sin(α)tan(α) ?

The answer to whether sec(α)-cos(α)=sin(α)tan(α) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove sec(α)-cos(α)=sin(α)tan(α)

prove

sec (α) - cos (α) = sin (α) tan (α)

True

Solution steps

sec (α) - cos (α) = sin (α) tan (α)

Manipulating left side

sec (α) - cos (α)

Express with sin, cos

- cos (α) + sec (α)

Use the basic trigonometric identity:

sec (x) = \frac{1}{cos ( x )}

= - cos (α) + \frac{1}{cos ( α )}

Simplify

- cos (α) + \frac{1}{cos ( α )} : \frac{- cos ^{2} ( α ) + 1}{cos ( α )}

- cos (α) + \frac{1}{cos ( α )}

Convert element to fraction:

cos (α) = \frac{cos ( α ) cos ( α )}{cos ( α )}

= - \frac{cos ( α ) cos ( α )}{cos ( α )} + \frac{1}{cos ( α )}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- cos ( α ) cos ( α ) + 1}{cos ( α )}

- cos (α) cos (α) + 1 = - cos^{2} (α) + 1

- cos (α) cos (α) + 1

cos (α) cos (α) = cos^{2} (α)

cos (α) cos (α)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cos (α) cos (α) = cos^{1 + 1} (α)

= cos^{1 + 1} (α)

Add the numbers:

1 + 1 = 2

= cos^{2} (α)

= - cos^{2} (α) + 1

= \frac{- cos ^{2} ( α ) + 1}{cos ( α )}

= \frac{- cos ^{2} ( α ) + 1}{cos ( α )}

= \frac{1 - cos ^{2} ( α )}{cos ( α )}

Manipulating right side

sin (α) tan (α)

Express with sin, cos

sin (α) tan (α)

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

= sin (α) \frac{sin ( α )}{cos ( α )}

Simplify

sin (α) \frac{sin ( α )}{cos ( α )} : \frac{sin ^{2} ( α )}{cos ( α )}

sin (α) \frac{sin ( α )}{cos ( α )}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{sin ( α ) sin ( α )}{cos ( α )}

sin (α) sin (α) = sin^{2} (α)

sin (α) sin (α)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

sin (α) sin (α) = sin^{1 + 1} (α)

= sin^{1 + 1} (α)

Add the numbers:

1 + 1 = 2

= sin^{2} (α)

= \frac{sin ^{2} ( α )}{cos ( α )}

= \frac{sin ^{2} ( α )}{cos ( α )}

= \frac{sin ^{2} ( α )}{cos ( α )}

Rewrite using trig identities

\frac{sin ^{2} ( α )}{cos ( α )}

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= \frac{1 - cos ^{2} ( α )}{cos ( α )}

= \frac{1 - cos ^{2} ( α )}{cos ( α )}

We showed that the two sides could take the same form

\Rightarrow True

csc (x) = \frac{1 + sec ( x )}{sin ( x ) + tan ( x )}

tan^{2} (θ) + 6 = sec^{2} (θ) + 5

\frac{cos ( x )}{1 + sin ( x )} + tan (x) = sec (x)

sec (\frac{π}{2} - x) = csc (x)

\frac{1}{csc ( x ) - 1} + \frac{1}{csc ( x ) + 1} = 2 tan (x) sec (x)

Is sec(α)-cos(α)=sin(α)tan(α) ?
The answer to whether sec(α)-cos(α)=sin(α)tan(α) is True